Term Rewriting System R:
[X, Y, L]
eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(s(X), s(Y)) -> EQ(X, Y)
INF(X) -> INF(s(X))
TAKE(s(X), cons(Y, L)) -> TAKE(X, L)
LENGTH(cons(X, L)) -> LENGTH(L)

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Size-Change Principle
       →DP Problem 2
MRR


Dependency Pair:

EQ(s(X), s(Y)) -> EQ(X, Y)


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





We number the DPs as follows:
  1. EQ(s(X), s(Y)) -> EQ(X, Y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.


   R
DPs
       →DP Problem 1
SCP
       →DP Problem 2
Modular Removal of Rules


Dependency Pair:

INF(X) -> INF(s(X))


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(INF(x1))=  x1  
  POL(s(x1))=  x1  

We have the following set D of usable symbols: {INF, s}
No Dependency Pairs can be deleted.
8 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
DPs
       →DP Problem 1
SCP
       →DP Problem 2
MRR
           →DP Problem 5
Non-Overlappingness Check


Dependency Pair:

INF(X) -> INF(s(X))


Rule:

none





R does not overlap into P. Moreover, R is locally confluent (all critical pairs are trivially joinable).Hence we can switch to innermost.
The transformation is resulting in one subcycle:


   R
DPs
       →DP Problem 1
SCP
       →DP Problem 2
MRR
           →DP Problem 5
NOC
             ...
               →DP Problem 6
Non Termination


Dependency Pair:

INF(X) -> INF(s(X))


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

INF(X) -> INF(s(X))

R = none

s = INF(X)
evaluates to t =INF(s(X))

Thus, s starts an infinite chain as s matches t.

Non-Termination of R could be shown.
Duration:
0:01 minutes